In class my professor said that $$ \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = -\frac{2\pi}{b}\sin(ab) $$ where $a,b > 0$. However, since the poles are on the real axis, isn't the integral equal to $$ \pi i\sum_{\text{real axis}}\text{Res}(f(z); z_j)\mbox{?} $$ If that is the case, the integral is $$ -\frac{\pi}{b}\sin(ab). $$

  • $\begingroup$ The more serious question is: Is this improper integral convergent? $\endgroup$ Nov 13, 2013 at 4:09
  • 5
    $\begingroup$ @TedShifrin The general mantra for this type of integral in complex analysis is to ignore that sort of convergence issue when you can get a contour limit that exists. At which point you just call it the "principal value" of the integral in question, rather than claiming it to be its value. Of course, one immediately abuses notation and writes things just as if it is equal to that value, anyway. $\endgroup$ Nov 13, 2013 at 5:22

4 Answers 4



$$f(z):=\frac{e^{iaz}}{z^2-b^2}\;,\;a,b\in\Bbb R^+\;,$$



$$\gamma_{r,s}:=\{r+se^{it}\;;\;0\le t\le \pi\}\;,\;r,s\in\Bbb R^+\;,\;\Gamma_R:=\{Re^{it}\;;\;0\le t\le \pi\}\;,\;\;R\in\Bbb R^+$$

Since $\;f(z)\;$ analytic on and within $\;C_R\;$ , we get


By the corollary to the lemma in the second answer here we get

$$\begin{align*}\lim_{\epsilon\to 0}\int\limits_{\gamma_{-b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=-b}=i\pi\frac{e^{-iab}}{-2b}=-\frac{\pi i}{2b}e^{-iab}\\ \lim_{\epsilon\to 0}\int\limits_{\gamma_{b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=b}=i\pi\frac{e^{iab}}{2b}=\frac{\pi i}{2b}e^{iab}\end{align*}$$

And applying Jordan's Lemma the integral on $\;\Gamma_R\;$ goes to zero when $\;R\to \infty\;$ , so in the end

$$0=\lim_{R\to\infty\,,\,\epsilon\to 0}\oint\limits_{C_R}f(z)dz=\int\limits_{-\infty}^\infty f(x)dx-\frac{\pi i}{2b}\left(e^{iab}-e^{-iab}\right)\implies$$

$$\implies \int\limits_{-\infty}^\infty f(x)dx=-\frac\pi b\frac{e^{iab}-e^{-iab}}{2i}=-\frac\pi b\sin(ab)$$

and I get the same as you did...


$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ Assumming the integral meaning is "principal value $\pp$ of it": $$ \pp\int_{-\infty}^{\infty}{\expo{\ic ax} \over x^{2} - b^{2}}\,\dd x = \int_{-\infty}^{\infty}\expo{\ic ax}\bracks{% {1 \over x^{2} - b^{2} + \ic 0^{+}} + \ic\pi\delta\pars{x^{2} - b^{2}}}\,\dd x \tag{1} $$

\begin{align} &{1 \over x^{2} - b^{2} + \ic 0^{+}} = {1 \over \pars{x - \root{b^{2} - \ic 0^{+}}}\pars{x + \root{b^{2} - \ic 0^{+}}}} \\[3mm]&= {1 \over \bracks{x - \pars{b - \ic 0^{+}}}\bracks{x - \pars{-b + \ic 0^{+}}}} \quad \mbox{since}\ b > 0.\tag{2} \\[3mm] &\delta\pars{x^{2} - b^{2}} = {\delta\pars{x + b} + \delta\pars{x - b} \over 2b} \quad \mbox{since}\ b > 0.\tag{3} \end{align} Since $a > 0$, we choose a "${\tt\mbox{half moon}}$" contour in the upper complex half plane:

We replace $\pars{2}$ and $\pars{3}$ in expression $\pars{1}$: \begin{align} &\color{#0000ff}{\large% \pp\int_{-\infty}^{\infty}{\expo{\ic ax} \over x^{2} - b^{2}}\,\dd x} = 2\pi\ic\,{\expo{\ic a\pars{-b + \ic 0^{+}}} \over \pars{-b + \ic 0^{+}} - \pars{b - \ic 0^{+}}} + {\ic \pi \over 2b}\,\bracks{\expo{\ic a\pars{-b}} + \expo{\ic a\pars{b}}} \\[3mm]&= -\,{\pi\ic\expo{-\ic ab} \over b} + {\ic \pi \over 2b}2\cos\pars{ab} = \bracks{-\,{\pi\ic \over b}\,\cos\pars{ab} - {\pi \over b}\,\sin\pars{ab}} + {\pi\ic \over b}\,\cos\pars{ab} \\[3mm]&= \color{#0000ff}{\large -\,{\pi \over b}\,\sin\pars{ab}} \end{align}


A related problem. Here is how you advance

i) Using partial fraction gives

$$ \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = \frac{1}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iax} }{ \left( x-b \right) }}dx-\frac{1}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iax} }{ \left( x+b \right) }}dx.$$

ii) Making change of variables $x-b=t$ and $x+b=t$ in the two integrals on the RHS yields

$$ \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = \frac{e^{iab}}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iat} }{ t }}dt - \frac{e^{-iab}}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iat} }{ t }}dt=\\ \frac{i\sin(ab)}{b}\int_{-\infty}^{\infty}{\frac {e^{iat} }{ t }}dt. $$

iii) the integral

$$\int_{-\infty}^{\infty}{\frac {e^{iat} }{ t }}dt =-i\pi$$

is well known and exists in the Cauchy principal value sense. Check Fourier transform too.

iv) To evaluate the last integral using residue theorem, consider the complex integral

$$ \int_C \frac{e^{iaz}}{z}dz, $$

where $C$ is the contour (for $a>0$)

enter image description here


This integral is unsolvable in real domain, because the singularities are not removable so integral should diverge.but you can take contours in complex plane which connect negative and positive infinitis and then you get the above result.


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